Answer:
The mean is 27.8
The required variance is 23.9358
Explanation:
Consider the provided information.
Here the random sample size, n = 200
Unemployment rate is 13.9% = p = 0.139
Part (a) expected number of grads who are unemployed.
Hence, the mean is 27.8
Part (b) variance of the random variable "number of grads who are unemployed.
![Variance = np(1-p)\\Variance = 200* 0.139(1-0.139)\\Variance = 27.8(0.861)\\Variance = 23.9358](https://img.qammunity.org/2020/formulas/mathematics/college/51xx1nrwvx2yzv580bee6tti45vp9kwqgx.png)
Hence, the required variance is 23.9358